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Here are some direct questions at the interface of algebraic and differential geometry:

(1) Is there an easy characterisation of those affine algebraic varieties which are Kahler?

(2) Is there an easy characterisation of those affine algebraic varieties which are symmetric spaces?

(3) Is there an easy characterisation of those affine algebraic varieties which are both? (From the first comment below, it seems that we can rephrase this question as: which affine algebraic varieties are symmetric?)

(4) What happens if I then also require compactness?

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Smooth affine varieties are always Kahler: restriction of Euclidean metric from $\mathbb{C}^n$. Smooth projective varieties are always Kahler: restriction of Fubini-Study. – Qfwfq Nov 11 '10 at 22:05
Some comments: (1) All Stein spaces (effectively: closed submanifolds of $\mathbb C^n$) are Kahler. I don't imagine the converse holds. (2) Elie Cartan classified all Riemannian symmetric spaces based on their homotopy groups. If I recall correctly, then having a complex structure is equivalent to having a holonomy group contained in $U(n)$. (3) I don't think so, see (1). (4) Same response for (2), extremely difficult for (1). In dimension 1, everything is Kahler. In dimension 2, having even 1st Betti number is equivalent to being Kahler. For dim > 2, this is a completely open question. – Gunnar Þór Magnússon Nov 11 '10 at 22:07
Compact affine algebraic varieties are unions of finitely many points, by the maximum modulus principle. – David Speyer Nov 11 '10 at 22:08
Oh, wait, you only want algebraic varieties? Then everything is Kahler as soon as it is smooth. As unknown said, affine varieties are Kahler because $\mathbb C^n$ is Kahler, and projective varieties are Kahler because $\mathbb P^n$ is Kahler. This becomes a difficult question when you consider arbitrary complex manifolds. – Gunnar Þór Magnússon Nov 11 '10 at 22:10
@Abtan: As a complex algebraic variety the sphere is not affine. – babubba Nov 11 '10 at 23:07
up vote 1 down vote accepted

If we ignore the trivial case of the affine line, then irreducible symmetric spaces come in pairs compact - non-compact. The compact ones are naturally projective varieties, while the non-compact ones are affine varieties. Thus question (4) is problematic, unless you mean "locally symmetric" or a more general notion of symmetric space than I understand here (i.e. "globally symmetric Riemannian symmetric space"). As far as non-compact symmetric spaces are concerned, they are Kähler if and only if they are biholomorphic to a bounded symmetric domain. Equivalently, there exists a compact quotient with non-trivial H^2 or, equivalently, the point stabilizer of the automorphism group has infinite center... I could give many more characterizations, but I do not quite see what you are after, so maybe you can provide more detailed information?

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Well if the compact irreducible symmetric varieties are the projective varieties, then since all projective variaties are Kahler, isn't the answer: compact irreducible symmetric Kahler varieties = projective variaties? – Abtan Massini Nov 11 '10 at 22:09
But projective varieties are not affine. You asked for affine varieties... Otherwise, the answer is, that compact symmetric spaces are K\"ahler if and only if they are complex projective. All compact symmetric spaces are real projective, but this does not imply K\"ahler. – Tobias Hartnick Nov 11 '10 at 22:31
Well the projective varieties are just a subset of affine varieties - with inclusion under realification (see this question Thus, I think you've given me what I asked for. – Abtan Massini Nov 11 '10 at 22:37
So, to be clear, you are saying that: compact symmetric Kahler varieties <---> complex projective varieties – Abtan Massini Nov 11 '10 at 22:50
No. Among the symmetric spaces, the compact Kähler ones are the same as the complex projective ones. But not every complex projective variety is a symmetric space! – Tobias Hartnick Nov 11 '10 at 23:14

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